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Acceleration command in proportional navigation?
I was recently taking a gander at proportional navigation, a missile guidance law, and I noticed most PN laws output something known as an acceleration command. The acceleration command is a command given to the fins of the missile for it to steer. The only problem being the command is in feet/s2, and not in an angular velocity like d/s.
So how would I go converting this acceleration command into an angle for the fins to rotate to? Would I need to just do a bit of testing and adjusting? How would the command rotate the missile? The acceleration is already in the direction the fins have to rotate to to steer the missile. Any ideas?
I also tried missile autopilots, but they seemed too complex for the task at hand. Is there any easy (or relatively easy) formula for the creation of a pseudo pilot that can convert the command into the rotation required for the missile (or for the ailerons) in a place without external forces (like air density, etc. etc.)

As a note, this is a repost from the physics stack exchange, as I was sent here.

This is between physics and aviation, so I would say that it'd be OK here and at the physics.SE.

Does eq. 45 help? This is an application of the lift equation, α is the angle of attack which for a fin with no camber is the angle of the fin.

Do you have a link to an example of what you describe?

Interesting, could you explain a bit eq. 45 as I dont really understand all of the terms.

Also, here is link to understand what Im saying. As you can see he uses a direction cosine matrix to apply the acceleration command to the body coordinate system. I need to understand how to turn that into angular velocity for my missile.  https://www.youtube.com/watch?v=yUlC6GZEUVs&list=PLcmbTy9X3gXt02z1wNy4KF5ui0tKxdQm7&index=8&t=783s&pp=2AGPBpACAYgEAQ%3D%3D

As a last note, I dont completely understand the terms pV2 or SC in equation 45

Hi! Thank you so much! This really helped. But lets say my missile doesnt generate lift (in my simulation aerodynamic forces do not apply), how would I calculate it then? Would it just be mF = alpha, or something different?

@KINGMOOSE Ah, I did not realize you were on the lookout for a simplified model. I updated my answer with such a simplified model

The control structure for a typical missile looks something like this:

As you can see, the PN-Guidance command is carried out by a dedicated acceleration controller. This has a couple of resons:

First an foremost, the missile has some pretty complicated flight dynamics. The varying airspeed and altitude of the missile means highly varying dynamic pressure acting on the missile. At high airspeeds, minimal fin deflections already mean very high accelerations, while at low airspeeds, the same fin deflections almost do nothing.
Most missiles are non-minimum phase systems due to the location of the fins at the rear of the missile. This means, when a fin is deflected, the acceleration will first be negative before becoming positive. This is a vicious effect, which the Autopilot is specifically designed to take care of.

The job of the acceleration controller is therefore to make the missile behave nicely. It can be designed such that the PN-Guidance always sees a relatively uniform missile behavior across all flight speeds, and altitudes.
Edit: After reading a couple of your comments, I would recommend the following: You can use what I wrote above to your advantage. Instead of explicitly simulating the acceleration controller in combination with the missile dynamics, you simply assume that the acceleration controller does its job right under all circumstances, and model the combination of these two systems as a low-pass of second order with a rise time of a few hundred mili-seconds. Additionally, you simply neglect the rotational dynamics (again, your assumption is that the controller does its job right). This leaves you with a modeled point mass, which follows lateral acceleration commands with the dynamics of a second order system. For some kind of realism, I would also include drag (perhaps the drag could be dependent on the g-forces the missile pulls). The resulting velocity and position can then be simply integrated from these accelerations. The resulting dynamic system would then be something along the lines of:
$$
\ddot{\vec{x}} = R_{IB} \cdot \begin{pmatrix} -a_D \\ a_y(t) \\ a_z(t) \end{pmatrix}
$$
with $R_{IB}$ being the rotation matrix rotating the accelerations from the body coordinates to inertial coordinates, $a_D$ modeling drag acceleration, $a_y(t)$ and $a_z(t)$ being the (commanded) acceleration of the PN-Guidance after it passed the second order dynamics.
Perhaps I can draw a sketch of the situation tomorrow if you are interested.

Hi there! This is super interesting! I would love a more detailed explanation, as unfortunately I work in a pretty basic game engine, and I dont know how much of this it can do on its own. If you could provide a sketch explaining the steps you go through that would be great!

If I remember correctly you can derive the angular rotation speed during a stationary turn from the acceleration command by just dividing by the current velocity:
q_rate = acceleration / speed
To compute the necessary fin angle you need to have a physical model of the missile in question and you need a control loop which adjusts the fin angle until the desired acceleration is measured with the accelerometer in the rocket.
For an analytical approximation you need to find a function that describes how much acceleration the missile can produce for a certain angle of attack and airspeed. This obviously depends on the size, mass and shape of the missile and on the thrust produced. These parameters will change significantly throughout the flight, because the mass of the rocket is changing and the thrust will stop once the rocket runs out of propellant during the coasting phase.
Once you know what angle of attack is needed you can compute the fin angle necessary to rotate the missile towards this target angle, based on the inertia of the body and the torque that the fins produce at the current speed per angle of deflection (probably linear up to moderate deflection angles). This relationship can also be approximated from basic aerodynamic principles like the lift formula, moment arm, etc.
So to summarize: There is no simple solution to your question. It depends on the type of missile and the aerodynamic forces it can produce. A control loop for the fin angle is probably needed to dynamically adjust to the current conditions.

Hi there! Can I possibly make an example? So after having an acceleration ( in my case a vector3) we divide that acceleration by our missiles velocity, and this gives us angular velocity (presumably in d/s). I cant place my head around why that would work? Could you maybe explain it further? And to respond to the rest of your answer, I am trying to replicate a simulation engagement in a game engine, and I have complete control over all aerodynamic forces and components. Is there any formulas that can make somewhat of a missile pseudo pilot for these conditions?

@KINGMOOSE It doesn't give it in deg/sec, it gives it in something proportional to deg/sec. That constant of proportionality depends mostly on the geometry (and mass) of the missile and slightly on some other stuff. If you're programming a sim, you can just pick that constant to be whatever works well

I see! Thanks balot! Ill try this system later today!

I also tried missile autopilots, but they seemed too complex for the task at hand.

They are exactly as complex as required for the task at hand.
Missile dynamics are extremely complicated. There are no simple formulas.

Yes of course, but I didnt think I would need them in my case where external forces (such as drag, air density etc.) dont apply.

I am attempting to simulate an engagement without these forces.

@KINGMOOSE Exactly what forces do you want to model if you're ignoring all external ones?

I am attempting to simulate a Proportional Navigation system for my missile. I only want to have the guidance system and an acceleration actuator that can turn the acceleration command in an angle (or angular velocit) command.

In basic I want to start with a missile which can perfectly turn given an acceleration. I just want to figure out how to convert acceleration into a desired rotation/angular velocity.


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Why disable stall warning based only on low airspeed, rather than multiple criteria?
One of the factors that potentially contributed to the Air France 447 crash is that the stall warning was disabled when the airspeed fell below 60 knots.
I understand why this is -- you don't want to have the stall warning go off while the plane is parked at the gate or slowly taxiing.
However, this seems like clearly bad design for two reasons (that should have been obvious to the designers):

This introduces an extra point of failure. If the pitot-static system fails, then the stall warning may be disabled due to erroneously low reported airspeed.
As demonstrated by AF447, a stalled plane may have very low forward airspeed in an extreme stall.

Moreover, it seems like the solution is easy: Only disable the stall warning if a number of criteria are met. For example, the following logic could be used.

The the stall warning (based on AoA vanes) is disabled if and only if (i) the forward airspeed is below 60 knots AND (ii) the landing gear is down/on the ground AND (iii) the radar altimeter reports a height of 0.

This logic would still prevent erroneous stall warnings on the ground, but it is more unlikely to suppress "real" stall warnings, as it would require multiple systems to fail.
Alternatively, if you are worried about one broken sensor resulting in erroneous stall warnings, you could require only two of the three conditions to hold for the stall warning to be disabled. That's still better than only one condition.
My question: Why do aircraft not use multiple criteria, such as suggested above, for disabling the stall warning?
The point of my question is that you want the stall warning disabled when the plane is on the ground. Whether or not the plane is on the ground should be rather obvious and you should not need to try guessing whether the plane is on the ground based on airspeed alone. There are plenty of other sources of information that can indicate whether the plane is on the ground.

Where does it say it's a contributing factor? It says that you need airflow over the AoA vanes, and if you're under 60 kts, the AoA reading is likely to be invalid, so there's no way to tell if you're in a stall (pg 44). I think the simple fact is that if you're in the air at 60 kts, recovery is highly unlikely.

Also, see fig 11 on page 44 and the note. It uses all three measured airspeeds to determine the stall threshold, which is dependent on airspeed. So the idea is to use triple redundancy to always have good data, instead of making up potentially dangerous invalid data.

@user71659 I’m suggesting more redundancy, not less. AF447shows how triple pitot tubes can all simultaneously fail - they all point forwards and have the same vulnerability to ice.

You're getting two different things confused. The stall system already takes the worst case mach number (pg 44 note). Increasing the reliability of airspeed instrumentation is a different question. The fact of the matter is, if airspeed is really under 60 kts, then you are dead, there's no reason to warn about stalls in that case.

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Is a restricted airworthiness certificate required to operate a Cessna O-2A Skymaster in part 137 operations?
Does a Cessna Skymaster O-2A need to be classified as "restricted" if it is to dispense chemicals over a congested area, in a part 137 operation? The O-2A we're currently flying is classified as standard and the category is normal. It is a surplus army aircraft if that makes any difference.

Will you be doing low level application? Are the chemicals toxic? One needs to look at the operational and aircraft adaptations needed for the operation you are proposing.  So tell us more about your operation and the equipment, and you may get a better answer to your question.

Using economic poisons?  Also, normally 137 may not operate in congested areas. Is this where your land is?

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Does bypass air produce thrust?
How does bypass air exiting the bypass nozzle produce thrust in a turbofan engine?
I thought that thrust was produced by air reacting with the fans. So if thrust happens when there is an action-reaction pair with the air and fan blades, is there additional thrust provided when the mass of air pushed back by the fan blades exits the back of the engine?

Possible duplicate of What provides the greatest thrust in a high-bypass turbofan engine?

That other thread does not really answer this question. I haven't been able to find a straight answer online, but am I correct in assuming that thrust in a high bypass turbofan engine is provided by the action-reaction of air and the fan blades (ignoring the thrust from the engine core) and NOT from the air exiting the bypass nozzle?

More likely a duplicate of Is there a pressure rise and velocity change across a turbofan fan and stator?

Also What is the bypass air in a turbofan engine actually for?

Yes a lot of the thrust is from the discharge of the fan being crammed into the convergent nozzle and forced to accelerate.  Similar to a regular jet engine where most of thrust is coming from the air coming out of the turbine being forced to accelerate by the convergent nozzle of the tail pipe.
Best way to envision it is to look at a balloon that you blow up and let go.  In that case all of thrust is coming from static air being forced to accelerate out the neck of the balloon by the rubber compressing it.

So if thrust happens when there is an action-reaction pair with the air and fan blades, is there additional thrust provided when the mass of air pushed back by the fan blades exits the back of the engine?

The thrust provided by the action of the fan blades against the air is the same thrust as the thrust provided by the air exiting the back of the engine. If an engine is producing 200 kN of thrust, then you will see 200 kN of force exerted by the fan blades against the air, and you will see 200 kN of exhaust (measured by its mass flow rate times its speed) exiting the back of the engine.

You can model the fan disc driving the bypass air flow as a well-shrouded propeller inside a short duct. It does indeed generate thrust which, as in the general case of a propeller, is transmitted to the engine case via a thrust bearing that supports the fan disc and thence to the airframe via the engine mounts.

So the air exiting the back of the bypass duct (which has now been accelerated due to the bypass nozzle) does not produce any forward thrust?

no, it produces lots of forward thrust.

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How can I calculate distance based on altitude and glideslope angle?
How do I calculate the distance to touchdown when the glideslope angle is different from 3°? For example, you are 670ft AGL on a 3.2° glideslope approach; what do I change in the formula to calculate the distance?

Which formula are you referring to?

How about doing it for 3° then subtracting 7% (a close approximation of .2)?

@Simon subtracting 7% is quite fair (see also my answer on approximation).

Just in case you don't have a calculator at hand (yeah, I know...):
WolframAlpha returns (unsurprisingly): 1.97nm (and lot of other entertaining info).

Sidenote: Because $\tan(\alpha)\approx\alpha~for~small~\alpha$, you can use a simple linear approximation in this case. Your new glide distance will be more or less $\frac{3}{3.2}=\frac{15}{16}=0.9375$ times the original one.

Or paste this into google.
670 ft/tan(3.2 degrees) to nautical miles

The $300ft/NM$ used to calculate your height above the runway at a particular distance is just an approximation that is easy to remember, and is close enough for our purposes.
If you want a more exact number, you can use some trigonometry:
$$\tan(3)=\frac{xft}{6,076ft/NM}=318ft/NM$$
(the "real" number).
For 3.2 degrees, it would be:
$$\tan(3.2)=\frac{xft}{6,076 ft/NM}=340ft/NM$$
So if you want to know the appropriate altitude for, say a 3 mile final on a 3.2 degree glideslope, it would be:
$$Runway~Touchdown~Elevation+340ft/NM\times3NM=TDE+1,020ft$$
In your specific example where you already know the altitude and want to know when to start the descent, the answer would be $2NM$:
$$\frac{670ft}{340ft/NM}=2.0NM$$
($1.97$ if you want to be REALLY exact.)

Since the 1 in 60 rule of thumb yields a solution of 300 feet for every NM from the touchdown point for a 3 degree slope, use 320 feet for a 3.2 degree slope.  So 670ft, gives you a nadgers whisker over 2 nm.
To be a little more precise, 2.09375 nm ($\frac{670}{320}$)

DeltaWhisky, not sure why your edit disappeared but I must point out that a nadgers whisker has a very precise meaning in general aviation.  A badgers whisker is quite different, so I've edited it back.

Just out of curiosity, how did you determine that a 3.2 degree slope needs a 320 ft/NM descent?  Is there a rule of thumb that you are using to determine that?

As you say. it's just an approximation but 300ft per NM at 3 degrees is good enough for GA.  It's an extension of the 1 in 60 rule. So, you can say 100ft per degree.  3.2 degrees is therefore 320 ft.

$$tan(3.2)=\frac{670}{a}$$
$$\frac{670}{\tan(3.2)}=a$$
$$a=11,983ft$$
$$11,983ft=1.97NM$$

Since we all agree that a 3.2°GS give us 340'/nm, I suggest that you take your FAA, subtract by the TCH elevatin and divide by 340, the result is the distance to touchdown

Thanks to all for answers. The $\frac{1.97}{96}$ was not correct.
This is how you should do it:
$$3.2=\frac{670ft\times60}{3.2}=12,562ft=2.066NM$$

The 1.97 nm is the precise value. The rule of thumb overestimates the value slightly.

As Jan says, 1.97 is the actual correct mathematical value.  Your answer does not accurately answer the question as you originally asked it....

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Are rolling takeoffs now more common than powering up while holding on the brakes?
I remember back in the 90's that commercial planes would line up on the runway, stop, apply full power and then release the brake to take off.
Now I've been on flight where they've literally rolled from the taxiway straight onto the runway and then powered up without stopping.
Why has that changed? What were the reasons for the older style?

Back then people were in less of a hurry.

the older style lets the engine ramp up to max thrust earlier on the runway (on the starting point even) and get more speed over a shorter distance.

@ratchetfreak Makes sense. So then why was it stopped?

@CameronMacFarland engines that ramp up faster and are more powerful which eliminates the need for a longer runway otherwise, and noise reduction

Whether to use a static or a rolling takeoff is usually at the discretion of the captain as long as performance is not an issue and ATC doesn't need you to wait on the runway.
Reasons to do a static takeoff include:

Less runway is required
Better obstacle clearance (mainly because of the earlier liftoff)
Takeoff engine power can be confirmed prior to brake release
ATC may require an aircraft to wait on the runway due to wake turbulence separation requirements
The aircraft manufacturer may not allow a rolling takeoff

Reasons to do a rolling takeoff:

There is sufficient runway available
Obstacles are not an issue
Takeoff engine power can be set relatively quickly and an abort can still be made at a low speed
They take less time to perform, and at busy airports this can mean getting another airplane or two out/in per hour
They offer more passenger comfort because of the smoother acceleration

Not all airplanes provide performance information for rolling take offs. In this case, a commonly used technique is to ensure that takeoff power is set by a certain point on the runway and adding that distance to your calculated takeoff roll for planning purposes.

your bullets are correct. I believe the really telling one is that they take less time. Time is $$. Having said that, I fly as a passenger out of LaGuardia frequently, and when using runway 13 it is about 50-50 that the pilot will use a classic short field technique: stop in position, brakes on, power up, brake release, and go.

FWIW, if the tower order was to line up and hold, we would do a static takeoff, although certainly not hold the breaks while takeoff power was being brought up unless performance required it. If the tower clearance was cleared for takeoff (without any line up and hold), we always did a rolling takeoff.

@Terry Good point, I have added line up and wait to my answer!

Though not common among modern airplanes, there have been a few aircraft where a static takeoff was a manufacturer requirement. Concorde comes to mind.

@BretCopeland Added, thanks!


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What routes are available for airliners to fly from Europe to Japan without entering Russian airspace?
Under the present circumstances it seems like for an European vector flying above Russian airspace can be dangerous (I don't know if Russian airspace is currently closed).
However reaching Japan from Europe requires flying above Russian territory.
With closed Russian air space, what are the options for a passenger route connecting, e.g. Schiphol with Narita?

https://www.cnn.com/travel/article/ukraine-conflict-world-air-map/index.html might be of interest.

The great-circle route from Kagoshima to Istanbul doesn't (quite) pass through Russian territory; Kagoshima-Athens is similar.

It's not that it's dangerous (Ukraine's is dangerous right now, with was confirmed by a lot of NOTAMs). It's that it's closed by Russia for a lot of countries (in response to their closure for Russian companies).

While it won't be the most efficient or direct routes, flights would be redirected South of Russian airspace.
They're probably avoiding Ukrainian air space too, just to be on the safe side. A quick browse of Eastern European airspace on FR24 confirms a large hole in tracked traffic over Ukraine at this time.
Thanks to tml's comment, here is a map image of the Lufthansa flight he noted.

Another option would be to go North across the North Pole instead of East. Based on this Great Circle map from Amsterdam to Narita, the GC distance would be 5805 miles. However, adding in Fairbanks to get it to go the other way increases the distance to 7720 miles.
I'm not sure what airports are along the way via the southern route to force a GC map down south, and there would probably be several options. I would imagine though, that a more southerly route is still going to be shorter.
Heading further south to avoid China, should that become necessary, I chose Ho Chi Minh City as a stop between Amsterdam/Schipol and Narita. That route ends up being 8848 miles and the GC route still crosses Western China, so the actual route would be even further, making the "fly East to go West" option a better one.
In summary:

AMS-NRT: 5805 miles
AMS-FAI-NRT: 7720 miles. An increase of 1915 miles or 33%.
AMS-SGN-NRT: 8848 miles. An increase of 3043 miles or 52%.

Comments are not for extended discussion; this conversation has been moved to chat.

Another option would be to fly over the pole to Anchorage (4450 miles = 7160 km) and from there to Tokyo (3466 miles = 5577 km).  This combined distance (plus a fuel stop) of just over 7900 miles (appr. 12,700 km) compares to the existing Great Circle route over Russia of 5540 miles (appr. 8915 km) -- "only" 50% longer.
I doubt the route referenced in another answer is much shorter than this, since it's well off Great Circle, but my route would require fueling arrangements in Anchorage that Lufthansa probably doesn't already have, vs. fueling in Kazakhstan (give or take a 'stan), perhaps even India, which they might already have.
Routing via Anchorage, however, also avoids flying over China, which could become a complication depending how things continue to play out with Russia.

Avoiding China was a thought I'd placed in a very geographically challenged (and, therefore self-deleted) comment on my answer. Having to do so will probably make the "via Alaska" route end up being shorter. Also, I'd imagine that Anchorage/Fairbanks will be more than happy to work out refueling contracts with all the major airlines.

They've done it on short notice for freight transporters (FedEx and UPS, etc.) in the past -- due to volcanic eruption, IIRC.  A little more complicated for passenger flights, where passengers and crew need "fuel" (and crew need rest) but not insurmountable.

Anchorage was a common technical stop for passenger flights to/from East Asia before Siberian overflights opened up after the fall of the USSR, and still is for freight.

I have called my local representative in Congress requesting they bring up the possibility of reopening Anchorage Airport to international travelers without a U.S. visa, just like how it was before 9/11.

@StephenS Current U.S. immigration law requires that everyone transiting the U.S., even on technical stops, have a valid visa and go through U.S. immigration and customs; they have no concept of international airside transit. Also, stopped at LAX, not stops in Hawaii: they no longer operate service to London, and when they did, it was via LAX. Before 9/11, those who were transiting could go without a visa and be let into the country by showing their onward ticket at immigration, and that's how transit at Anchorage worked back in the Soviet era. My call was to reinstate this arrangement.

@gparyani I’ll delete the comment, but that’s not how technical stops are supposed to work.

@StephenS That's true, but the U.S. toughened up its law in response to 9/11 and it's never been changed due to the usual bureaucratic delays.

@gparyani I know the US has always been stupid about transit connections, but I didn’t realize we also broke the case of passengers not even getting off their plane.

@StephenS Not saying this policy makes sense, but passengers and crew would likely need to be allowed off the plane, at least into the terminal, for a stop like this between 7-8 hour legs...

@ZeissIkon they might like to get off the plane at Anchorage but given those who fly on Emirates Auckland-Dubai service are on the plane for a single 17 hour leg it might be considered optional ;-)


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How does a turbojet engine's performance change as the load on it varies?
What happens when the load increases on the turbojet engine? What senses the speed change in jet engines and how speed correction takes place?

This seems to be a re-post of this now deleted question. Again, please add more details to your question! The engine itself doesn't do anything to correct for a change in load (the pilot might, or the electronics might). Please describe the exact situation you are asking about.

Ahh, a governer!.

@RobertDiGiovanni- Yes tell me something about the governor.

A governor would make more sense on a turboshaft. For a turbojet the pilot selects the power which will result in a certain amount of fuel "calculated" by the fuel control system (mechanical and or hydraulic in the past to account for variables like ambient conditions, nowadays from software logic/engine control) which in effect determines the speed of the aircraft (equilibrium drag vs. thrust). How would you define increase of the load (load variation in the title)? Power off-take? Or adding more fuel? Please make this question more clear! We'd be happy to answer.

@user61081  Proper foundation would be application.  Turbines have a net output.  Now, what is the load?  They can be directly loaded with a power shaft (attached to a generator, fan, or propeller), or they may be simply pushing out thrust against the drag and friction load of the aircraft.  What is the application?

Taking a very common turbofan as an example, the CFM56-5B used on Airbus A320 family.

Indeed the pilot doesn't adjust the throttle when the power demand changes, e.g. when the anti-ice system is turned on, or when the angle air enters the engine varies with aircraft attitude, or when a larger electric power is needed for the galley, or when rain starts falling. All that is regulated.
The regulation is done by a computer in the engine itself, the FADEC (full authority digital engine controller). The crew or the auto-throttle sets a thrust demand, that's the parameter which is important for the flight, and the FADEC manages the engine in order to provide the power to deliver this thrust and the power for secondary uses like electricity.
The first parameter controlled by the FADEC is the fuel flow, the main way to determine how much energy is converted into heat and ultimately into air flow acceleration and propulsive force. But that's not the only parameter under control.
From the engine documentation:

The FADEC provides automatic engine thrust control and thrust parameter limit computation. The FADEC manages power according to two thrust modes:

manual mode depending on Throttle Lever Angle (TLA),
autothrust mode depending on autothrust function generated by the Auto Flight System (AFS).

This regulation is done using several means: fuel flow, variable vanes angle of attack in compressor, discharge valves, and turbine blades clearance regulation when temperature changes.

Image from CFM56-5B Technical Training Manual by Airbus
The FADEC manages secondary power demand variation like for anti-icing, cabin pressurization and air conditioning.

The FADEC also provides two idle mode selections: minimum idle and approach idle, obtained when the slats are extended. The idle can also be modulated up to approach idle depending on: air conditioning demand, wing anti-ice demand, engine anti-ice demand and oil temperature (for Integrated Drive Generator (IDG) cooling).

Thrust cannot be measured in flight, it must be evaluated from other parameters. Two usual parameters are:

N1, the rotation speed of the low pressure spool,

The engine pressure ratio (EPR), that is the pressure ratio between engine exit and inlet nozzle.

These two parameters, associated with other data, allow the FADEC to adjust the fuel flow and to monitor the engine efficiency. How they are exactly used depends on the engine design.
The FADEC is the blue box in the image below, receiving many inputs (top right) as temperatures and pressures and controlling regulation elements shown on the engine drawing.

Image from CFM56-5B Technical Training Manual by Airbus
A FADEC or an equivalent controller is required on turbine engines due to the complexity and the extreme temperature of the combustion, pushing the combustion chamber and the turbine materials to their limit, so fuel must be regulated very precisely.
A regulation is also needed to obtain the best engine efficiency, and to not exceed engine limitations of many sorts. Before FADEC use, monitoring and controlling the engines was the task a flight engineer, the third crew member:

Flight engineer panel on a Boeing 747F, source
FADEC, source Safran:

Other engines may not have a full authority controller, that is a computer which cannot be overridden by the crew, but a similar controller exists, generally named an EEC (electronic engine controller).

Additional information
You can see one aspect of this regulation in this video. Valves open and close in order to prevent the compressor to stall when thrust demand is changed.
Also see these questions:

Why does the CF6/CFM56 Turbofan Fan Frame contain 12 circular holes between the primary and secondary flow? Here.

How does the FADEC prevent compressor stall during rapidly increasing thrust demand? Here.

What is the difference between variable and transient bleed valves? Here.

Because turbines feed air to themselves with a compressor, they generally operate with variable rpm to control power output.
Adding more fuel without changing rpm (airflow) risks overpressuring or overheating the combustion chamber.  Opening up the jet exhaust nozzle more helps compensate for the delta Volume caused by delta Temperature. Variable stator vanes help fine tune engine output on the inlet side.
Jets operate using way more air than they need for stoichimetric combustion, so there is some room there for adding more fuel at the same rpm. Turbine helicopter rotor systems are of interest.


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Which materials were used in the F-100 Super Sabre construction?
I was wondering if anyone could help me with any details about the materials percentage on the F-100 Super Sabre. Considering the plane is quite popular even to this day, I'm a bit frustrated that I cannot find any info regarding this on the web.


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How can I import Garmin Pilot checklist files into ForeFlight?
Inside the airplane that I am training on, there is an iPad running the Garmin Pilot app with all the checklists for that plane in the app as a "Checklist Binder". I use my own iPad (a second iPad) on a kneeboard with ForeFlight for my sectional charts, flight planning, weather, TFRs, NOTAMs, PIREPs, AIRMETs, METARs, and everything else.
I was able to export the checklist binder from Garmin Pilot as a .gplts file from the iPad (the one that lives in the plane I am renting) and email it to myself. On my iPad, I was able to download the Garmin Pilot app and import the checklist binder really easily.
I'd like to export/import these checklists (this checklist binder) into my ForeFlight app so that all of my relevant information can be in the same place. Does anyone know a straightforward way to do this?
Currently, I have ForeFlight and Garmin Pilot on my iPad and I have to switch between them, which is a lot better than having to sit one iPad on top of the other when I need to go through a checklist. But I'd really like to be able to move these checklists into ForeFlight and keep everything in the same application.

Just a thought: Garmin Pilot does show weather, TFRs, etc., so you could just switch to it.

I posed a similar question to ForeFlight in May 2021 (how to import my text checklists to FF) and there reply was that import is not supported - here is their reply (and I never received any follow up indicating they were considering it further)

Currently, there is not a way to way to import checklists at this time. The only workaround that you could do in the meantime is upload your checklist to documents as a pdf. It won't be saved in the checklists section but would be in your EFB.

We appreciate the feature request though and I will pass your request on to our Product Management team for their consideration.

Bummer... Hopefully, they get to it soon, won't hold my breath. Thanks for your answer!

@currenthandle I have added support for the ForeFlight format to the excellent free EFIS Editor from rdamazio. I don't have a licence for Garmin Pilot and have no idea how it handles checklists. Maybe it's format can be implemented as well and then you have a full converter.


text:
How to compute the direction of the vector perpendicular to the wing?
I'm trying to compute the direction of the vector that extends perpendicularly up from the top of the aircraft's surface.
Is there a way to compute the vector's 3-dimensional parameters in Cartesian dimensions?
I have available the values of roll angle, pitch angle, and heading angle (relative to north).

To start with, you need to define your coordinate system. What are the axes defined relative to, and in which directions do they point?

@ThomasPerry: the question as it is now better reflects what the asker actually wanted to ask and the answers were update consequently.

@ThomasPerry: please read the comment of the asker self which I linked in my previous comment, that explains everything.

@sophit  No problem.

You cannot calculate the direction of the lift vector from roll, pitch and heading. The total aerodynamic force $\vec{F}$ is by convention split into lift $\vec{L}$ and drag $\vec{D}$ as follows:

drag is the force component parallel to the direction of relative motion,
lift is the force component perpendicular to the direction of relative motion.

(Wikipedia - Aerodynamic Force, Wikipedia: image)
In order to split the force, you need to know the direction of relative motion, i.e. the velocity vector $\vec{v}$. With this you can calculate lift and drag:
$$ \vec{D} = \frac{\vec{F} \cdot \vec{v}}{|\vec{v}|^2} \vec{v} \; , \qquad
\vec{L} = \vec{F} - \vec{D} $$

If you are just interested in the unit $\vec{Z}$ vector of the aircraft in the world coordinate system, this is given by
$$ \vec{Z} = \left( \begin{matrix} \sin(\phi) \sin(\psi) + \cos(\phi) \sin(\theta) \cos(\psi) \\
\cos(\phi) \sin(\theta) \sin(\psi) - \sin(\phi) \cos(\psi) \\
\cos(\phi) \cos(\theta) \end{matrix} \right) $$
with $\phi$, $\theta$ and $\psi$ as roll, pitch and yaw respectively (see also Euler Angles and Rotation Matrices and What is the relation between roll angle and pitch angle?).

Thank you for the reply. I realized that the question is titled incorrectly. I meant to ask how to obtain the direction of a vector that is normal to the aircraft, not specifically the lift vector.

@james_erikson Ah, that has nothing to do with the lift vector. I added the vector calculation based on roll, pitch and yaw.

Thank you very much. If my coordinate system were: up (+z), north (+y) and east (+x), I would have to re-derive Z correct?

@james_erikson Have a look at the picture in this answer. In your coordinate system, you'd basically have to swap roll and pitch and then use -Z instead of Z.

I'm using a simplified dynamics model that doesn't take into account the aircraft's yaw in the body frame. I only have heading in the inertial frame. Would that mean I simply remove the rotation matrix around the Z axis? Or I'm thinking I would apply the yaw matrix before any other rotations?

@james_erikson Your heading is the yaw angle

@james_erikson - We use a down-pointing z-axis because we count headings clockwise. If you use a z-axis that points up, you get counterclockwise headings, i.e. west = 90°. That's inconsistent with standard aviation conventions.

Mr. Erikson, WHY do you want to know the direction perpendicular to the surface of the airframe? That is of no use, for anything. It is certainly not in any way relevant to the totally arbitrary definition of the force vector we call LIFT. That is defined as the component of aerodynamic force that lies normal (perpindicular), to the aircrafts flight path, not to the aircraft itself. We define it that way to separate or isolate the effect of aerodynamic force on the direction of aircraft velocity from its effect on the magnitude of the velocity, (speed).

If your question is really about LIFT, then an approximate (ignoring slight lateral shifts due to yaw), answer is just to add 90 degrees to the aircrafts velocity vector, in the plane of defined by the aircrafts roll axis and yaw axis.

You can convert any vector between body coordinates (x=forward, y=right, z=belly) and navigation coordinates (x=north, y=east, z=down) by multiplying it with the rotation matrix. For the inverse conversion, multiply the vector with the transposed rotation matrix.
By your problem description, you have a vector that points "up" in body coordinates and you want it in navigation coordinates, like this:
\begin{align}
\mathbf{v}_{body} &= \begin{bmatrix} 0\\ 0\\ -1\end{bmatrix} \\
\mathbf{v}_{navi} &= \mathbf{A} * \mathbf{v}_{body}
\end{align}
Construct the rotation matrix from Tait-Bryan angles (ϕ=roll, θ=pitch, ψ=heading) using:
\begin{align}\mathbf{A} = \mathbf{A}_Z\mathbf{A}_Y\mathbf{A}_X
\end{align}
with:
\begin{align}
 \mathbf{A}_X &= \begin{bmatrix} 1 & 0 & 0\\ 0 & \cos\phi & -\sin\phi\\ 0 & \sin\phi & \cos\phi \end{bmatrix} \\[5px]
 \mathbf{A}_Y &= \begin{bmatrix} \cos\theta & 0 & \sin\theta\\ 0 & 1 & 0\\ -\sin\theta & 0 & \cos\theta \end{bmatrix} \\[5px]
 \mathbf{A}_Z &= \begin{bmatrix} \cos\psi & -\sin\psi & 0\\ \sin\psi & \cos\psi & 0\\ 0 & 0 & 1 \end{bmatrix}
\end{align}

Thank you, this would be for a North-East-Down coordinate system?

@james_erikson - Yes, the result is a vector in north-east-down coordinates.


text:
What are the mechanical deflection angles for Airbus side-stick controllers?
Does anyone know the angular deflection limits of the longitudinal (pitch) and lateral (roll) axis of Airbus's side-stick?

From this site:

So max deflection in pitch is +/- 16 deg, in roll it is +/- 20 deg

There is a rather quick YouTube video by a gentleman called Capt. Joe which can partially answer your question.

The limits of the effective deflection of the sidestick is presented on the PFD, which that also changes with many variable factors to allow the aircraft to operate within its approved and safe performance envelope.

What's presented on the PFD is elevator & aileron deflection, not sidestick deflection.

And anyway the answer should be in the text, not in the linked video only (or it should be a comment).

I tried to put in the link in my comment, but unfortunately the whole  YouTube piece popped up as my answer. Sorry for disturbing you, I beg your forgiveness.


text:
Would removing heat from the compressor and adding heat to the turbine increase a gas turbine's efficiency?
Since a compressor compresses air, the air temperature increases due to P1/T1=P2/T2. When the temperature is increased it would be harder to compress - so would removing the heat from the air inside the compressor reduce the energy required by the compressor, and increase the efficiency of the engine? Also, since the turbine accepts hot air and uses the expansion of the gas to do work, would adding the heat extracted from the compressor to the turbine area increase efficiency even further?

This is really a thermal dynamic problem and you will get a much better answer at physics.se

If you find an efficient way to extract heat from a lower temperature area & add it to a higher temperature area, without using a lot of energy in the process, you'll have a valuable invention on your hands!  Welcome to Av.SE, by the way.

Heat removal like that is sometimes done in gas turbine engines, not by cooling the compressor during operation, but by the use of an isobaric intercooler between the low and high pressure compressors.  These kinds of configurations are common in industrial gas turbines used for electric power generation, marine propulsion and other applications where extra weight is not as large of an issue as is with aviation gas turbines.
As far as adding this extracted heat energy back into the gas flow through the turbine section, this would not be possible as compressor outlet temps are in the neighborhood of 450°F where exhaust gases exit at around 1000°F, preventing a return of the compressor heat to the turbine section for extra power.  Typically intercooler exhaust is simply vented into the atmosphere.
More commonly, gas turbines are used in combined cycle powerplants which make use of the gas turbine exhaust heat to power a water boiler driving a steam turbine and secondary generator.  Some of the thermodynamic cycles used in these systems can be quite elaborate and can achieve an exergetic efficiency of over 60%, making them the most efficient heat engines on the planet.

TL;DR: You are absolutely right!
your basic considerations are correct. There can be a thermodynamic advantage (inter)cooling the compressor and pre-/re-heating the turbine.
The obstacles are in the engineering and commercial field.

Engineering: Designing heat-exchangers which are small, light, and low-loss.
Commercial: Reduce cost.

I am currently only aware of one european research program (NEWAC to overcome these obstacles but I am sure there are others.
The following illustration is from NEWAC’s homepage:

It is possible to increase efficiency of a gas turbine in two ways: by cooling air before it enters a compressor, or by heating air after it leaves a compressor. It is therefore possible with an internal heat exchanger -  if the heat flows opposite to what is posed in the OP: from the hot exhaust gases to the air delivered by the compressor.
From the Wikipedia article for Brayton cycle (which is what takes place inside a gas turbine):

Recuperator[14] – If the Brayton cycle is run at a low pressure ratio and a high temperature increase in the combustion chamber, the exhaust gas (after the last turbine stage) might still be hotter than the compressed inlet gas (after the last compression stage but before the combustor). In that case, a heat exchanger can be used to transfer thermal energy from the exhaust to the already compressed gas, before it enters the combustion chamber. The thermal energy transferred is effectively reused, thus increasing efficiency as well.

Note that the heat must be added after the compressor/before the combustion chamber. From the same article:

Transferring heat from the outlet (after the last turbine) to the inlet (before the first compressor stage) would reduce efficiency, as hotter inlet air means more volume, thus more work for the compressor.

In that light, cooling air before it enters a compressor increases efficiency. Adding heat after the turbine decreases efficiency but does increase thrust, as afterburners demonstrate.

What about connecting the recuperator to (inlet and outlet of) the combustion chamber so that the compressor outlet air is heated by the turbine inlet air? This would lower the turbine inlet temperature to compressor exhaust temperature. Would this help for better fuel economy (less fuel for the same thrust)?

@Kozuch The turbine extracts energy from the hot gas flow. Cooling the gas flow before it reaches the turbine results in less energy available for maintaining the Brayton cycle. Best to extract the heat after all work is done - behind the turbine, form the exhaust gases.


text:
Private aircraft seating and exits - any rules?
Airliners and charter jets, operating under Part 121, 135, or their foreign equivalents, fall under regulations to ensure accessible exits, 16g seats, and other aspects of occupant safety.
Private aircraft, not holding out, don't need to follow these rules and can have extravagant interior layouts. I'm interested in which regulations or guidelines is the interior designer still expected to follow.
Can the plane be filled with bar stools, rocking chairs, etc., or are there provisions under which something absurd, like a swimming pool on a 737 BBJ, would be denied an airworthiness certificate?

It seems like you’re talking more about the composition of the seating and not the actual amount of seats, as well as amenities. For that you need to look at CFR § 25.853 I would additionally look at AC 25.853-1 which covers flammability testing. Basically even just refitting the existing seats with a nice leather material may involve expensive testing of the material to ensure it doesn’t burn. Also anything wood is basically out.
The whole sub part D would probably be worth looking at. It includes things like:

§ 25.601 General.
The airplane may not have design features or details that experience
has shown to be hazardous or unreliable. The suitability of each
questionable design detail and part must be established by tests.
§ 25.603 Materials.
The suitability and durability of materials used for parts, the
failure of which could adversely affect safety, must—
(a) Be established on the basis of experience or tests;
(b) Conform to approved specifications (such as industry or military
specifications, or Technical Standard Orders) that ensure their having
the strength and other properties assumed in the design data; and
(c) Take into account the effects of environmental conditions, such as
temperature and humidity, expected in service.

Which I think would be a non starter for the ideas provided, at least in the United States.

Thanks. Yes, I'm mostly looking for what regs the seats on a Part 91 bizjet have to comply with. How to quickly tell complete non-starters from something that needs testing.

My answer applies to aircraft standards so it wouldn’t matter if it’s your own plane or American Airlines they would need to meet these standards.

Emergency exit and seating design is controlled by the aircraft certification process, not through the operating certificate. Operators may not exceed the certified seating capacity. Part 25 covers larger commercial aircraft including the BBJ and similar. Part 23 covers general aviation. Here are excerpts:
Part 25

14CFR25.807g Type and number required.  The maximum number of passenger seats permitted depends on the type and number of exits installed in each side of the fuselage. Except as further restricted in paragraphs (g)(1) through (g)(9) of this section, the maximum number of passenger seats permitted for each exit of a specific type installed in each side of the fuselage is as follows: Type A 110, Type B 75, Type C 55, Type I 45, Type II 40, Type III 35, Type IV 9.

Exit type in Part 25 is specified by door size and location.
Part 23

§ 23.2315 Means of egress and emergency exits.
(a) With the cabin configured for takeoff or landing, the airplane is designed to:
(1) Facilitate rapid and safe evacuation of the airplane in conditions likely to occur following an emergency landing, excluding ditching for level 1, level 2, and single-engine level 3 airplanes.
(2) Have means of egress (openings, exits, or emergency exits), that can be readily located and opened from the inside and outside. The means of opening must be simple and obvious and marked inside and outside the airplane.
(3) Have easy access to emergency exits when present.
(b) Airplanes approved for aerobatics must have a means to egress the airplane in flight.

Part 23 was rewritten to be objective oriented rather than regulated by specific design. The means of compliance now contains all the specifications that used to be in the regulation. The original regulation may be used in many cases but alternate means of compliance can also be proposed and used if approved. Getting approval for a means of compliance is not trivial, so it can be easier just to use the original reg.

Thanks. So I guess capacity is locked. Any idea if anything applies to what the seats can be, or is it just the total count?

@Therac There is a maximum capacity for any given certification but you can put as few seats as you like. There are several rules about width of space between rows and seat clearance at the exits. There are separate certification rules for seats, as for side facing 'couch' type seats and upper body restraints to keep heads from bashing into things. A Supplemental Type Certificate (STC) is used to create a permanent approval for a particular modification to all listed aircraft and form 337 is used to get approval for a particular modification to one aircraft. The lines are not always clear.

@Therac There is a distinction between a seat used for takeoff and landing and some other piece of furniture. Your barstool might be ok for inflight use (no turbulence) but would not be allowed for TO/L.


text:
How does the F-16 side stick convert pilot input to an electrical signal?
In What is the motivation behind designing a control stick that does not move?
the F-16's control stick is discussed. What is the method that is used to sense the pilot's input to the stick and convert this to an electrical signal?

It's 2-dimensional set up with load cells, like this one.

pretty much like how all the scales works now.

The stick measures force by using strain gauges. These work by measuring the change in electrical resistance caused by stretching or compressing the thin metallic foil, which is bonded onto the stick shaft.
An accurate setup requires two strain gauges at opposite sides, which are measured using a Wheatstone bridge: the difference in resistance of the two gauges is used for the force signal.
There are other ways of measuring input force, but none are more accurate than the strain gauge/ Wheatstone bridge setup which can measure force in both directions and has little to no hysteresis.

Does the stick move?

@rclocher3 yes it does, a little bit, loaded by a very stiff passive spring. Key point though is that it isn’t stick position that is used for input to the fly-by-wire computations, but stick force is. There is a breakout position in the centre where the stick does not move until a force level is exceeded, no problem for fly-by-wire if the force is measured and the position is inconsequential.

The side stick controller uses two sets of quad-redundant, linear variable-displacement transducers to measure stick position against a linear spring force. One set is for pitch and the other is for roll. The electrical signals from the LVDTs go to a computer. Control law software interprets the inputs and controls the hydraulic actuators that move the control surfaces.


text:
Why are the spoilers used as the backup controls instead of ailerons on the Boeing 777?
While reading this question, I ran across an interesting unanswered comment.  Why did Boeing choose to use some of the spoilers as mechanical backup controls rather than the ailerons?

The B777 is a high subsonic aeroplane, and at high airspeeds the ailerons remain stationary because wing twist may cause aileron reversal.
By using inboard spoilers as mechanical backup roll control surfaces, there is roll control available at high airspeeds as well. Roll authority would be limited at approach speeds due to the smaller moment arm than the aileron provide and due to negative lift on one side only.

By transitory measure, is this meant to serve the time between dual engine failure (fuel starvation or APU off) and RAT deployed? The batteries cannot tie over this transient?

@Jimmy Cannot find an exact reference and have amended. It looks like a last resort measure.

Re item 2 - the spoilers are also hydraulically-powered on the 777. The 'mechanical' does not refer to actuation, but to signalling the surface to move.

@RAC Right-o, thx

May involve a new question, but what failure mode is this addressing (referring to "last resort measure"?

If you're going fast enough for twist-induced control reversal to be an issue, aren't you already going way too fast?  (Also, why wouldn't the asymmetric drag and negative lift produced by roll-spoiler deployment be just as apt to twist the wings and cause a roll reversal?)


text:
Does a high "wing loading" require a high approach speed or does it cause a high approach speed?
Raymer says:

Landing distance is largely determined by wing loading. Wing loading affects the approach speed, which must be a certain multiple of stall speed (1.3 for civil aircraft, 1.2 for military aircraft). Approach speed determines the touchdown speed, which in turn defines the kinetic energy which must be dissipated to bring the aircraft to a halt. The kinetic energy, and hence the stopping distance, varies as the square of the touchdown speed.

My question is : In which manner does wing loading affect approach speed?
I'll explain my question better:

The first option:
Higher wing loading implies less wing surface. Therefore, you have a higher value of stall speed (because, for the same lift needed and the same CL and the same density, with a small wing surface area, you have to increase speed to compensate for less surface), so you need a higher value of approach speed.
In this case : a higher wing loading REQUIRES a higher approach speed.
The second option:
Higher wing loading causes less wing surface. Therefore, you have less exposed surface, so you have less drag, so it’s easier to maintain a higher speed.
So if this last sentence is correct (I’m not sure), in this case it’s not only the required speed, but the CAUSED speed.

What do you think?

SOURCE:   DANIEL P: RAYMER “Aircraft Design: A Conceptual Approach”, Chapter 5 “THRUST-TO-WEIGHT RATIO AND WING LOADING”, Paragraph 5.3 WING LOADING, Subparagraph: “Landing distance”

It's the first of your options.
The effect of reduced drag with reduced wing area is dominant only at high speed, when friction drag is dominant. At low speed induced drag is the major source of drag, and with a given planform a smaller wing will create more drag at the same speed, not less.
Due to the possibility of wind shear or gusts you should keep some margin to your stall speed during approach, right down to the flare. Therefore, the safe approach speed dominates the landing distance, as Raymer writes.

The lift equation: $L = C_L \cdot ½ \rho V^2 \cdot S$, with S the wing area. Wing loading is defined as L/S, equalling: $$\frac{L}{S} = C_L \cdot ½ \rho V^2$$.
With a given $C_{Lmax}$, a higher wing loading implies a higher speed. The first option is the correct one.


text:
Why was the boat mounted this way on the underside of the wing on the "Landseaire" flying yacht?
After seeing the below image of the "Landseaire" flying yacht here I was curious to know why the small boat mounted to the underside of the plane was mounted the way it was.  At least to me (and @Dave who mentioned it in his answer containing this picture), I found it interesting that the boat would be mounted "backwards" and also so far from the body of the aircraft.  Can someone explain why this position and orientation was necessary or why it was chosen?

If you want to read more about the story of this specific aircraft read that. Unfortunately it doesn't give any reasons about why the boat is mounted like that, although one photo seems to suggest that the contour of the boat matched the wing closer in that configuration.

I doubt it was due to the mounting points on the wing and the boat, and it was simply easier to mount it that way. It may not be a performance / aerodynamic decision.

My first thought was that it keeps the rate of change in the cross section smooth, as it appears right behind the engine.  Then again, I highly doubt this aircraft was worried about supersonic performance =)

It is also odd they did not attach something aerodynamic to the boat's flat surface to lessen drag and improve airflow. Or maybe they did but it is not pictured.

The shape of the aft part of a body is aerodynamically more important than the shape of the forward part of a body.

The reason is the "flat tail" of the boat. The drag this "flat tail" produces when mounted "reverse" is much less then the dynamic drag it would induce the other way around.
https://en.wikipedia.org/wiki/Drag_(physics)

I don't see how the link helps explain why the flat side produces less drag than the pointed side.– Ron Beyer

The published cw coefficients (see answer of @jwzumwalt ) show the best value for the drop shape facing towards the air stream.
There is no entry for a reversed drop. But it is evident that is would have a higher cw coefficient.
A closer look at the boat reveals it also has a drop shape. But this is facing to its tail. Therefore the boat creates less drag when mounted "reverse".

If this was the case, wouldn't the boat cause less drag moving backwards through the water also? – andy-m

It might but the usually draft of the boat is no so deep that a relevant part of the "flat tail" is actually in the water.

source: https://upload.wikimedia.org/wikipedia/commons/2/2d/Bundesarchiv_B_145_Bild-F056332-0004%2C_Bonn%2C_Bundesgartenschau%2C_Seen.jpg
On the other hand a little turbulences at the boats tail could help to keep direction in the absence of  a steering paddel...

This explanation is wrong - an anemometer spins with the forward facing blunt half of the hemisphere causing the most drag - en.wikipedia.org/wiki/Anemometer – jwzumwalt

The most important differences between the anemometer an the drag experiment is:
The half bowls in the anemometer are hollow which "catches" the wind way better than the flat surface of the filled half bowls in the experiment.

I don't see how the link helps explain why the flat side produces less drag than the pointed side. It could be that this is correct, however it seems more like a statement than any kind of explanation as to why...

If this was the case, wouldn't the boat cause less drag moving backwards through the water also?

@jjack based off of my rough estimate it is not but it is certainly close to the trailing edge.  Maybe 6 inches or a foot?

If the way the boat is fixed is more aerodynamic than the other way, then you should explain this to Nasa. They believe a prism with the flat face ahead generates 4x the drag of a bullet (though the Re number is not mentioned) :-)

@mins If drag has nothing to do with it, why don't we have any actual air plane with a "flat tail"? And for the resource you linked: the dynamic drag (and the shapes cw value) depends on the air speed...

This is interesting.  You cite a demonstration which shows something rather peculiar.  jwzumwalt shows a chart of Cd for various object which shows the more intuitive answer.  How exciting!  Two answers that disagree, AND cite their sources =)

This explanation is wrong - an anemometer spins with the forward facing blunt half of the hemisphere causing the most drag - https://en.wikipedia.org/wiki/Anemometer

@CortAmmon This example show that the cw coefficient of card mainly depends on the form of theit tail: https://en.wikipedia.org/wiki/Drag_coefficient#Practical_example I found a German resource telling that 80% of the cars drag is generated at its tail. And I strongly beleve that air crafts follow the same physics... ;o)

A cars drag in the forward motion is irrelevant to the discussion. What we are interested in is a comparison, you would need to show what the drag is if the car is driven backwards too. The wind tunnel test I documented shows Cd for both hollow (small dash lines) and flat (solid line) surfaces. SOLID flat surface into wind Cd 1.17, trailing SOLID flat surface Cd .42 - wind tunnels don't lie.

@jwzumwalt "wind tunnels don't lie." agreed. But cw coefficient depends on the airspeed too. The dymaic drag generated at the shapes tail raises quadratic with the speed while the static drag at the front raises linear.  So what air speed was used to find the cv coefficients?

@CortAmmon: Fluid dynamics (including aerodynamics) are always dependent on the flow regime. In particular, neither case mentions the Reynolds number (aka Re number, as mins calls it).

Note that boats designed for speed don't have blunt sterns -- racing shells are almost symmetrical with the rudder arrangements accounting for most of the difference.

With regards to the lesser question as to why it is mounted so far from the body, in this case it's actually the only available place; the Catalina is amphibious so mounting it on the fuselage/hull would cause issues with off-water operations. The internal structure of the wing also needs to be considered: the boat can only be attached were there is a structural hard point, and some method of releasing the boat. In this case, the boat is mounted where the underwing stores would normally be fitted, so there's already a release mechanism and internal supporting structure:
underwing.jpg
As for the primary question, my phd is in hypersonic design so I'm a little fuzzy on subsonics, but although mounting the boat backwards increases frontal drag, having the sharp prow facing aft reduces flow separation.
flow separation reduces pressure drag
This would help to reduce wake turbulence and reduce buffeting over the tail. Possibly it also minimises disturbances in the prop wash, since the boat is mounted so close to the engine. In this case, although the flat front isn't optimal for drag, the whole boat is shaped more like a tear drop fairing, with a sharper trailing edge. Without wind tunnel data or cfd analysis to back up this theory, it is only my surmising: as others have stated, it could simply be the contour of the boat matched the wing better that way.

Any reason why they couldn't have made the entire boat to be a little more aerodynamic or put in an aerodynamic shell?

@dalearn yes, possibly. But with the bureaucracy that goes into military decisions when it comes to design changes, getting a completely new boat design certified would have been a hassle. They could have just as easily attached a disposable fairing to the boat to make it more aerodynamic, if it was really an issue. Custom lifeboats for asr work definitely did exist. The B-17 was used for asr with a custom boat. link

The attachment of the boat as seen in the picture to reduce drag might seem to be common sense. However truth is sometimes stranger than fiction. We are given the drag of various 2d and 3d objects on page 3-17 of "Fluid-Dynamic-Drag" by Hoerner.
Both the 2d and 3d objects with the blunt end towards the rear - just like the boat moves through water - has the least drag. The drag in the water is going to closely approximate the drag in the air, and the boat should have been pointing into the wind to have the least drag.
In the 3d drag profile the #2 & 3 illustrations have CD of ~.40 while the opposite direction (#8 & 9, blunt first) has a Cd ~1.20, or about three times as much.
If the intent was to reduce drag, the designers or mechanics who installed the boat without wind tunnel data got it wrong! But, it is also possible the boat is situated in that direction due to the prow being higher and fitting the wing camber better. The boat does appear to match the camber quite well. Note the curve of the boat rail in the other picture.
We can look at a real world example and see that the wind tunnel data is correct. Hoerner in the same book provides Cd for various "boat tail" bullet shapes and they also have the least drag with the blunt end trailing. Bullets could use a different shape than we are accustom to, but tests show a bullet with a sharp point and blunt rear have the least Cd.
In fact, the Bell X-1 was "a 'bullet with wings', its shape closely resembling a Browning .50-caliber (12.7 mm) machine gun bullet" - https://en.wikipedia.org/wiki/Bell_X-1
An anemometer spins because the forward facing blunt half of the hemisphere has the most drag, meaning the boat is in the wrong direction for the least drag - https://en.wikipedia.org/wiki/Anemometer
Note the triangular shape with its sharp corners is also counter-intuitive to what we would normally expect.
I think the drag chart accurately depicts the boats drag. What appears to be causing a misunderstanding is we normally find a forward facing blunt blended body to have the least drag, however in this instance the boat has sharp corners that can "trip" the otherwise smooth flow. The sharp corners make this an exception to the rule as depicted in the drag chart.

Aerodynamics is very different at 108 knots of Catalina's cruise speed, 832 knots of Bell X-1's max speed and 1730 knots of bullet leaving the muzzle of M2 machine gun. The first case is subsonic, where the trailing edge matters a lot, while the later two are supersonic, where separation always occurs after the thickest point anyway and most important is sharp point to minimize contact with the shock wave. Also your image does not really show anything really resembling the boat.

You are grossly misinterpreting the chart. Streamlined fittings, wheel pants, etc are installed rounded end forward and tapered end back. The boat approximates that shape and is installed in the orientation matching everything else of similar shape.

@JanHudec I'd say that the boat kinda resembles the 'streamlined' image

I agree that this does not convincingly imply the boat would better be oriented forwards, still +1 for some actual wind-tunnel numbers.

As for why is at that position, the answer seems simple- there is where the hardpoints for bombs and torpedoes are- so mounting the dinghies that makes sense, both from structural and maintenance point. From the linked article:

Slung under each wing, where bombs and torpedoes used to hang, are two 14-ft dinghies.

As for the orientation, it looks like it was selected so that the boat fits snugly against the underside of the wing. It is repeatedly alluded to in multiple articles:
Here

Each boat fits snugly against the wing and is raised or lowered by a built-in electric hoist.

and here

The Landseaire had 14-foot dinghies under each wing, hoisted to fit flush by cables that had once lifted torpedoes and bombs ...

Note that the photo in question shows a product demonstration and looks like boats were attached to the wing in both orientations- bow first

Landseaire with the boat fitted stern first; image from flightglobal archives
and also stern first:

N68740 at Ontario, California in April 1957 as a "Landseaire"executive air yacht conversion;image from Ed Coates Collection

Interesting indeed that the boat in your pictures is closer to what i would expect (e.g. snug fit to wing, more aerodynamic shape, etc) while the boat in my picture looks like a rowboat or similar.

The boat you are claiming as bow first is actually stern first, as evidenced both by the hull shape and the nameplate position.

Fundamentally this question is related to Reynolds number and dynamic similarity.
In order for the flow about two similar geometrical shapes to be similar and thus for their drag coefficient $C_D$ to be similar both the respective flows' Reynolds numbers $Re$ and Mach numbers $M$ must be the same. Mach number is related to compressibility and for the case at hand (subsonic aircraft and boat) compressibility effects are negligible, so lets focus on $Re$.
Reynolds number is essentially the ratio of inertial to viscous or friction forces present in a flow: $$Re=\frac{\text{Inertial forces}}{\text{Viscous/friction forces}} \: .$$
Reynolds number is a function of the body (in this case boat) length, the speed of the flow and the viscosity (thickness/stickiness) of the fluid.
Lets estimate $Re$ for the boat in the water and mounted to the aircraft wing:
Assuming the boat is about $3m$ long and moves in the water at $2m/s$, and assuming the aircraft cruises at $200$ km/h at $10\:000$ ft, we can estimate $Re$ using this handy calculator and the fluid properties of water and air found here (you could also just do it by hand, its an easy formula).
It is found that:
$$
\begin{align}
Re_{\rm water} &\approx 6 \:000\: 000\\
Re_{\rm air} &\approx 10\: 000\: 000
\end{align}
$$
These values for $Re$ are of the same order of magnitude and the flow is in fact nearly dynamically similar. Note that $Re$ values typically span many orders of magnitude and these values are actually quite close.
Based on this, the same streamlining and drag reduction principles should hold and they would probably have been better off placing the boat the other way, i.e. the same orientation as it is made to be streamline in water.
It might however be that they simply placed it that way because it was easier to mount, as opposed to any aerodynamic reasons.
PS: I was half expecting the $Re$ of the boat mounted to the aircraft to be a lot higher than in the water, in which case I would have made the point that if the $Re$ of a body immersed in a flow is vastly different, the same streamlining principles would not hold, although in this case it would seem that they do.

Boats made to travel rapidly or efficiently in displacement mode don't have squared off transoms - traditional human powered boats have tapered sterns, as do naval vessels and (at the waterline) most keeled sailboats.  A squared off transom on a rowboat is more a manufacturing simplification to give a broad passenger area and convenience for mounting an outboard motor, in a hull that would be planing anyway if it were powered to move at any real speed.  For that matter on rowboats sometimes the bow is squared off, too.

Interesting, thank you!

Note that even if the Reynolds numbers are similar, the situations are quite different because the boat sits on the surface of the water, rather than immersed in it. In particular, this means wave drag is important even at low speed (basically, you have high mach numbers WRT surface waves, because these travel so slowly). Thus optimising the stern is little use – the bow will first be the bottleneck, adressed by https://en.wikipedia.org/wiki/Bulbous_bow.

Interesting @leftaroundabout, now that you mention that it seems obvious, thank you. For interests stake, is wave drag for aircraft and for boats caused by the same physical mechanism?

Broadly speaking, yes, but there are a couple of important differences. Water waves propagate in the 2D plane and have a nonlinear dispersion relation, which is why there isn't such a hard subsonic/supersonic distinction as there is for aircraft.

Wave drag has more in common with supersonic aerodynamics, where the compressibility effects and shockwaves play a big role. The water surface is compressible in the vertical direction.

If the boat was the other way round, it may produce some lift of its own, causing the wing to be pulled downwards, or may have an effect of the lift produced by the upper surface of the wing, refusing lift in that way. Making the airflow turbulent under the wing reduces the tendency for the pressure to drop.

+1 That's an interesting idea (really) however while it would prevent a rolling tendency, it now creates a yawing tendency (which could maybe be balanced by an asymmetric thrust setting).

This boat is subject to sub-sonic aerodynamics in the region where compressibility effects are negligible. In this region, streamlined bodies are teardrop shaped with a round front providing forces and a tapered rear to avoid separation bubbles and concomitant drag.
The boat has no rounded ends. So put its tapered end aft, to exploit one of the two properties of subsonic streamlined bodies.
And the stern has partially a rounded shape, which helps with streamlining when in front, not in rear. The flat surface does not help either in front or in rear.
So without knowing the exact Cd from wind tunnel measurements or CFD, the boat likely has least resistance with the bow pointing backwards.

The boat and the wing can be considered together, as one aerodynamic form.  The stern (of the boat) is in a region where wing AOA slows the relative airstream the most.
As "doppel wingers" know, the rear of the underside of the wing is where the airstream from the underside of the wing recombines with the down wash of the upper wing.  The flow is much faster and more turbulent here.  Trailing edge air flow properties were studied a century ago and applied to STOL aircraft as Junkers flaps.
For this particular boat/wing combination, the way they have it mounted would produce the least amount of drag.

Others have answered that the boats are so far outboard because they hang from the hardpoints designed for ordnance.
The reason the hardpoints themselves are so far outboard is twofold:  the ordnance is clear of the propwash to reduce drag, and when it drops, it won't hit the struts.  Both reasons apply to the boats, too.

One wouldn't think they'd be dropping the boats like they would have dropped ordnance. However, if one had to bail out... :)

Sure, "lowered gently."  But it still has to clear the struts!


text:
What are satellite time, GPS time, and UTC time?
What is the difference between Satellite time, GPS time and UTC time? How are these calculated from one another?

Both GPS time and UTC are derived from the atomic time TAI, a reference used to define the time unit in SI, kept by 400 atomic clocks in the world. GPS time unit and UTC time unit are the TAI second. In addition GPS time and UTC were equal on January 6 1980. End of 2021 the relationship is this one:

While the TAI scale is invariant, the year duration varies, mostly due to Earth decreasing rotation rate under the braking effect of tides. As humans, we expect noon of some day (Sun at its zenith), e.g. on February 19, to really happen each year at the same legal time. So the scale for legal time has to be flexible, unlike TAI. We use UTC for this purpose$^{(1)}$. UTC is TAI with leap seconds. Solar noon is delayed each year by some tiny amount, a mean 0.75 second. UTC matching with solar time relies on two actions:

Earth rotation period measurement based on the direction of remote quasars sensed by very large interferometers, done by IERS. This defines another time scale, UT1, which second is not atomic: Any UT1 day has exactly 24 hours of 60 minutes of 60 seconds.
UTC adjustment when the difference with UT1 is close to 0.9s by adding an atomic leap second to UTC. Due to Earth complex motion, a negative adjustment should be required in a near future. This never happened in the past, we don't know what could be the actual consequences. UTC is adjusted the last day of June or the last day of December. This day lasts one second more than an ordinary day, and 23:59:60 UTC actually exists.

Clock displaying a leap second in timezone UTC+9, source

In practice, few clocks or applications behave this way, most ignore the leap second, and most which do not use it correctly (e.g. some will slice the additional duration and distribute it over several hours before and after midnight).
The consequence of adding a leap second is the next UTC day lags behind the corresponding TAI day by one additional second and repeating this operation UTC differs from TAI by an increasing number of seconds. If TAI was convertible to a date using a calendar, we would have a difference in the date around midnight. But a calendar is only a notion valid for UTC, neither TAI nor GPS time are convertible to a date, they are only expressed in their canonical form, a count of seconds since the origin (epoch), a precaution which also protects them against bissextile years complexity.
While UTC time is adjusted for Sun zenith at constant time of the day, UTC date is similarly adjusted by an additional days each 4 years or so, to prevent the vernal equinox and Spring date to lag more and more. The reason is 365 days is not enough for Earth to complete its revolution around the Sun, it needs about 6 more hours. This adjustment maintains the Spring equinox in the range 19 to 21 March.
GPS time is never resynchronized; its offset from TAI is the number of leap seconds which existed on January 6 1980, that is 19. GPS and TAI are uniform (continuous) times; UTC is discontinuous due to leap seconds.
Note none of the "times" we measure today is an absolute time (e.g. since the Big Bang), they are a duration since some conventional close landmark (e.g. TAI was reset in 1958, UTC in 1972, GPS time in 2019). The duration expressed in seconds is the same as long as the duration unit is the same. This is the case for TAI, GPS and UTC which all use the atomic second defined in the SI. things get complicated only when converting this duration into a date/time, using the specific starting point and unequal years, days, minutes and hours introduced by leap years and seconds, which needs to be inserted properly. UTC is a broker of uneasy compromises, using the invariable atomic second unit, but kept in sync with UT1 which uses the variable solar second unit.
TAI:

Synchronized on existing UT2 on January 1 1958.

UTC:

Redefined in 1970 to include leap seconds.
Resynchronized on January 1st 1972 to be exactly 10 seconds behind TAI.
27 leap seconds have been introduced since then.
UTC is now 37 seconds behind TAI.
UTC = TAI - 37s at present.

GPS:

Defined as equal to UTC at midnight on January 6th 1980 when UTC was 19 seconds behind TAI.
18 leap seconds were added to UTC since then.
GPS time is now 18 seconds ahead of UTC.
The internal representation of GPS time was reset (due to week counter finite size) on August 21 1999 and April 7 2019, and this may happen every 1024 weeks. Conversion programs have to take this into account. While the start time is changed each time, the offset with TAI is not modified.
GPS time = TAI - 19s (always).
GPS time = UTC + 18s at present.

Time in space and time references for space
Time aboard a satellite, is a bit different than on Earth due to relativistic effects and altitude. For example on a Navstar satellite (GPS), time is slowed down by velocity and accelerated by the lower gravity in altitude, giving a global accelerating factor of about $5.3 \times 10^{-10}$:

If not accounted for, in one day it could build up to a timing error that would translate into a navigational error of 13.7 km.

(source)
There are time references for the deep space which are neither based on a rotating Earth nor on Earth gravity, e.g. geocentric coordinate time (TCG) is the time coordinate for geocentric celestial reference system (GCRS) and barycentric coordinate time (TCB) the time for barycentric celestial reference system (BCRS). Time elapses faster than on Earth, from Wikipedia:

Because the reference frame for TCG is not rotating with the surface of the Earth and not in the gravitational potential of the Earth, TCG ticks faster than clocks on the surface of the Earth by a factor of about 7.0 × 10−10 (about 22 milliseconds per year).

$^{(1)}$: The use of UTC instead of TAI, while usually presented like I did as a need to keep in sync with Sun, is actually based on an imbroglio mixing legal reasons and administration poor understanding of the huge risks and difficulties created by leap seconds for a little practical benefit. WRC 2023 conference will decide whether leap seconds should continue to be used for another 8 years or die. The clock is ticking...

Update on 10/2024: At WRC-23 it was agreed to increase the allowed difference UT1-UTC, to prevent UTC leap second discontinuities for at least a century. A proposal must now be drafted for an application before 2035, according to the already existing resolution 4 of CGPM-22. Instead of 0.9s, the difference could be as much as 100s. The leap second is not yet dead, but its life expectancy is no more than 10 years.

FYI, the often quoted claim of a 13km/day error is not correct. GPS location is computed by comparing the times of the several GPS signals, not comparing any of them to an earth based clock. Because all are at similar orbits all will be similarly affected, relativistic errors would be small, small enough not to be a deal breaker for the original purposes of GPS if they weren't accounted for.

@whatsisname: As the relativistic drift is time cumulative, there is a difference in how drift progresses on each individual satellite if orbits are not exactly circular with the same radius. I guess this difference is sufficient to create pseudo-range errors after a certain time. But I have no figures, I'm interested in related documentation.

https://physics.stackexchange.com/a/128951/63383

@Tim: Thanks. The answer you point to is based on the idea the pseudo-ranges are not computed from actual travel times ("It doesn't measure how long the signal took to get from satellite A to GPS"), and therefore there is no need to know the time of arrival, only the (common) departure time. But its not correct. The 4 signals don't have to leave the satellites at the same time and the travel times are computed and the receiver clock error estimated (both at once), This is a regular multi-lateration without a central clock.

Small but important correction: GMT does not exist anymore. You correctly note that GMT and UT1 are similar ideas, but only the latter is still rigorously defined, or even measured. Using GMT for modern timekeeping is an anachronism (albeit and regrettably, an extremely common one).

@imallett: Actually this is worse, GMT is wrongly but officially used by UK to refer to UTC. "when the clocks are 1 hour ahead is called British Summer Time (BST) [...] When the clocks go back, the UK is on Greenwich Mean Time (GMT)", source gov.uk. Indeed this is not GMT (UT1), but UTC with its leap seconds and constant second TAI scale.

Indeed. In many especially EU countries GMT is legislated as the standard; nevermind that it doesn't exist! AFAICT, the continued use of GMT is down to ignorance or British imperialist nationalism; all modern timekeeping— whether people know it or not —is done through UTC.

To clarify/reiterate my suggestion, though: UT1 and GMT are not "compatible" the way this answer currently implies. UT1 is not the new formal name for GMT; GMT literally doesn't exist, since it isn't measured anymore—in fact, it cannot be, as 0° longitude doesn't even pass through the Greenwich observatory in modern times. Whereas, UT1 is derived from interferometric measurements of distant quasars—i.e., the result is computed from sidereal days; it isn't directly related to solar time, even though it's expressed that way.

Equivocating UT1 ≈ UT ≈ GMT isn't completely bonkers, but it's technically incorrect because GMT doesn't exist today and because the notions are subtly incompatible besides. It would be better to talk about UT1 only. If you must mention GMT, mention also that it's obsolete (and therefore also don't explain the primary concepts using it). (anywho sorry for textwall and rant; HTH!)

Using GMT as an equivalent to UTC is in fact bonkers, the GMT date change was at noon in Greenwich, because it was an astronomy tool.
And then GMT (and UT0) fluctuate with wobbles in the Earth so they change every day.  Greenwich Civil time(GCT) was the practical precurser to UTC which is UT1 synchronized and stabilized to TAI seconds.

GPS time is the exact number of seconds since January 6, 1980, at 00:00:00 UTC. Some systems may (incorrectly) represent that as a date and time by assuming days of exactly 86,400 seconds.
UTC time truly has two parts: the date and the time within that date. Most days are 86,400 seconds long, but they may occasionally be 86,401 or 86,399 seconds long (due to “leap seconds”) to keep UTC in sync with Earth’s slightly irregular rotational period. Some systems may (incorrectly) represent that as a count of seconds.
GPS ephemera helpfully includes the net number of leap seconds added or subtracted under UTC, which is +18 as of 2021, so you can easily derive the current UTC date and time if that’s what you actually need.
Other satellite systems may have their own definitions of time.

Hi can you please elaborate more .
Let say satellite transmit 10:41:30 as time then how would the GPS time be calculated

@SourabhJain the satellite won't "transmit 10:41:30", but rather they will transmit "the exact number of seconds since January 1, 1980, at 00:00:00 UTC", as the answer says. That's GPS time.

note: GPS time is a linear count of SI seconds as observed on the surface of Earth. On the satellites, the clock tick a little slow to compensate for relativistic time dilation, which is (IIRC) on the order of 10¯⁶ while the clock have to be accurate to at least 10¯¹⁰ (more because the synchronization isn't every second) to provide the navigation accuracy they have.

Of course, defining a day as exactly 86400 seconds basically means that it has to be the seconds that vary in length, at least sometimes... Which is exactly what I understand Google did and which at least some NTP servers support. Also, obligatory xkcd: https://xkcd.com/2266/

@ilkkachu The “smearing” of leap seconds is pretty common these days because too much software breaks when time goes backwards under dumber handling of leap seconds. If POSIX time were TAI rather than UTC, it would be a lot simpler for everyone.

The first statement is wrong; the GPS epoch is 1980-01-06, not 1980-01-01. The 6th was the first Sunday of 1980.

@hobbs Fixed, thanks!

Disclaimer: Since I am using Wikipedia references below, this answer itself should not be used as a separate source for a any wiki article. Please look at the respective articles for the direct references to ensure they are the most current, accurate, and provide the proper source references.
When I was doing some Supplemental reading this question prompted I found this archived article from InsideGNSS, which talks about how time is compared between the GPS and Galileo, but also talks about the way the GPS System time is related to UTC. I added emphasis for the specific for this question but included the section for context. If you want more information you should probably review the article on the main site which may have more up to date information since none of the various systems are completely static.

Galileo runs on Galileo system time (GST), which is generated on the ground at the Galileo Control Center in Fucino, Italy, by the Precise Timing Facility (PTF), based on averages of different atomic clocks. GPS system time is computed by the GPS control segment using an assembly of clocks — the GPS Composite Clock, which consists of all operational monitor station and satellite frequency standards. GPS system time, in turn, is referenced to the U.S. Naval Observatory’s Master Clock and steered to Coordinated Universal Time — UTC(USNO), corrected for leap seconds (which GPS time doesn't use).

These two internal times are derived independently of one another but are kept close to the world’s reference time, coordinated universal time or UTC, with the offset between the two being precisely calculated on a continuous basis for the Galileo system by the PTF.

It would appear that at least these two GNSS providers reference back to UTC. UTC itself is derived from TAI (International Atomic Time), which is itself based on a weighted average of 400 atomic clocks spread around the world.
I did see one thing in reading that TAI article that I do not have the experience to resolve.  It appears TAI currently uses GPS system time to validate the timings between different atomic clocks. This would seem to lead to a cyclic Self Referencing error so I can only hope that was some how mitigated.
It could be that the nature of an encyclopedia entry is necessarily simplified leading to the apparent confusion. For practical purposes the time difference are probably extremely small.

Common-view GPS time transfer measures the difference between two clocks on the ground by referencing them both to GPS time. If GPS itself is fast or slow it doesn't matter, because that cancels out in the math. So there's no circular problem.


text:
How do pilots in the US identify the circuit (pattern) height?
In the UK, aerodrome weather information includes not only QNH (the air pressure at mean sea level), but also QFE, the air pressure at runway level. To arrive at an aerodrome, normally one obtains and sets the altimeter to QFE, so that it reads height above aerodrome level. Then if the circuit (traffic pattern) height is 1000 ft (which it usually is), you just read 1000 ft on the altimeter all the way around the circuit; for a standard overhead join it says 2000 ft; and so on. At any aerodrome the altimeter reads the same thing, because you always set QFE.
I've heard that QFE is not used in the US. Do you just have to work out the circuit height in your head by adding 1000 ft to the published aerodrome level? Or are circuit heights given amsl?

@Jim I didn't find the linked question incredibly helpful, because the question and answers are written with a lot of assumed knowledge about US procedures and terminology. In addition, one of the upvoted answers to this question would make no sense at all on the other question.

The questions are essentially the same.  It does not matter that the answers are different.  Different people will not necessarily answer identically.

My point isn't that the answers are different: it's that a good answer to one question doesn't answer the other question at all. I think that means the questions can't be duplicates, even though they're similar.

If you are referring to SentryRaven's answer the information he provided is irrelevant to both questions.  FAA sectional charts do not depict traffic patterns.

The Airport/Facility Directory (AFD) has that info. It lists the Traffic Pattern Altitude (TPA) in both MSL and AGL, as well as circuit direction.
QFE is not used in the US. So yes, you have to make the correction for field elevation. For example, an airport at 1200 MSL would have a pattern altitude of 2200 ft. (1000 AGL for props. 1500 AGL for Jets) and would be flown using that altitude on the altimeter.

Thanks for confirming that. Which of the two cases at the end of my question is true? i.e. does the runway information include a "circuit altitude amsl" entry, or do you just do the addition yourself when planning the flight?

You add the addition yourself.

The Airport/Facility Directory (AFD) has that info. It lists the TPA in both MSL and AGL.  As well as circuit direction.

The same basically applies to Europe. QFE is mostly used by glider pilots, whereas powered aircraft use QNH settings. Most visual approach charts or airport charts have the pattern altitude indicated at the visual representation of the pattern, at airports where no pattern altitude is provided or depicted, you use the rule of thumb to add 1000ft to the airport elevation.
Below image shows the pattern altitude at Cottbus-Drewitz, where the airport elevation is 274ft but the pattern altitude is 1100ft MSL and the other image the pattern at Lauenbrück, where the airport elevation is 76ft, but no pattern altitude is given. The calculated pattern altitude would be 1100ft (76ft rounded to 100ft + 1000ft).

(Image Source: www.flugplatz-drewitz.de - Pattern with Altitude in MSL)

(Image Source: www.edhu.de - Pattern without Altitude - Airport ELEV 98ft)

Sorry, I meant to respond to the OP, not you're comment.

-1 this does not answer the OP's question.

authors: محمد الحاج محمد